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Tuesday, 15 October 2013

Constructing simple two qubit states

Here, we will describe how to construct some simple two-qubit states using some primitive gates.

In what follows, a two-qubit quantum circuit consisting of one $CNOT$ gate and two Hadamard $H$ gates that computes the following unitary transformation will be described:
\[U=\begin{pmatrix}
1&0&0&0 \\
0&1&0&0\\
0&0&1&0 \\
0&0&0&-1
\end{pmatrix}\]

Observe that $U$ leaves the first three basis states unchanged and only multiplies the last by a factor of $-1$:
\[U\ket{00}=\ket{00},
U\ket{01}=\ket{01},
U\ket{10}=\ket{10},
U\ket{11}=-\ket{11}.\]

Consider the following circuit consisting of a Hadamard gate acting on the second qubit, followed by a controlled-NOT gate, and then with a second Hadamard gate acting on the second qubit again.

The action of this circuit on each of the four basis states is described below:
\[\ket{00} \overset{I\otimes H}\mapsto\super{\ket{00}+\ket{01}}\overset{CNOT}\mapsto\super{\ket{00}+\ket{01}}\overset{I\otimes H}\mapsto\frac{1}{2}(\ket{00}+\ket{01}+\ket{00}-\ket{01})=\ket{00},\]

The matrix representation of the gate $c_1H_2$ can be constructed by placing each of these output vectors as the columns of the matrix:
\[c_1H_2=\begin{pmatrix}
1&0&0&0\\
0&1&0&0\\
0&0&\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}\\
0&0&\frac{1}{\sqrt{2}}&\frac{-1}{\sqrt{2}}
\end{pmatrix}.\]

Then, matrix representation of the gate $c_1H_2$ can be constructed by placing each of these output vectors as the columns of the matrix:
\[c_1H_2=\begin{pmatrix}
1&0&0&0\\
0&\frac{1}{\sqrt{2}}&0&\frac{1}{\sqrt{2}}\\
0&0&1&0\\
0&\frac{1}{\sqrt{2}}&0&\frac{-1}{\sqrt{2}}
\end{pmatrix}.\]

Denote the following gate by $c_1S_2$:
First note that $S\ket{0}=\ket{0}$ and $S\ket{1}=i\ket{1}$. Now observe how the $c_1S_2$ gate acts on the four basis states:
\[\ket{00}=\begin{pmatrix} 1\\0\\0\\0\end{pmatrix}\overset{c_1S_2}\mapsto\ket{00}=\begin{pmatrix} 1\\0\\0\\0\end{pmatrix}, \ \ \ \ \ \ \
\ket{01}=\begin{pmatrix} 0\\1\\0\\0\end{pmatrix}\overset{c_1S_2}\mapsto\ket{01}=\begin{pmatrix} 0\\1\\0\\0\end{pmatrix},\]

The matrix representation of the gate $c_1S_2$ can be constructed by placing each of these output vectors as the columns of the matrix:
\[c_1S_2=\begin{pmatrix}
1&0&0&0\\
0&1&0&0\\
0&0&1&0\\
0&0&0&i
\end{pmatrix}.\]

Note that the action of $c_2S_1$ is identical to that of $c_1S_2$ from case (iii), and so the matrix representation of $c_2S_1$ is also the same:
\[c_2S_1=\begin{pmatrix}
1&0&0&0\\
0&1&0&0\\
0&0&1&0\\
0&0&0&i
\end{pmatrix}.\]

Here, the $4\times4$ matrix corresponding to the following quantum circuit will be given:
where $S$ is defined as it was above, and the last two-qubit gate denoted by $SWAP$ transposes the two qubits (more precisely, the $SWAP$ gate maps $\ket{00}\mapsto\ket{00}, \ket{01}\mapsto\ket{10}, \ket{10}\mapsto\ket{01},\ \text{and}\ \ket{11}\mapsto\ket{11}$).

Denote the overall action of this circuit by the unitary operator $U$. In order to express the action of $U$ as a matrix, first observe how $U$ acts on the four computational basis states:

Then the matrix constructed that contains the four output vectors as its columns gives the matrix representation of the unitary operation $U$ given by the circuit:
\[U=\frac{1}{2}\begin{pmatrix}
1&1&1&1\\
1&i&-1&-i\\
1&-1&1&-1\\
1&-i&-1&i
\end{pmatrix}.\]